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Movement along the cardioid ##r=k(1+cos\theta)##

  1. Aug 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moves with velocity ##v## along the cardioid ##r=k(1+cos\theta)##. Find ##\vec{r''}\cdot \hat{r}## and interpret the result. At what angles will this coincide with the centripetal acceleration?

    2. Relevant equations


    3. The attempt at a solution
    To find ##\vec{r''}\cdot \hat{r}## I will simply use the formula for the radial acceleration in cylindrical coordinates. This is ##\rho''-\rho\theta'^2## which after lengthy manipulation gives: ##-k(\theta'' sin\theta + \theta'^2cos\theta)-k\theta'^2(1+cos\theta)##. Which I have no idea how to "interpret", haha.

    To solve the second part well I know the position vector is ##\vec{r}=r\hat{\rho}## and that its derivative with respect to ##\theta## will give a tangent vector to the curve. This is simply ##\frac{d\vec{r}}{d\theta}=\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta}##

    The unit tangent vector is simply: ##\hat{t}=\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta})##. Which helps me to find the orthogonal normal vector and thus the centripetal acceleration as ##\frac{v^2}{R}\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\theta}-r\hat{\rho})##. Note I've switched components on the and sign on one component as well.

    This helps me notice that the centripetal acceleration coincides with my above result when the centripetal acceleration is purely in direction of ##\rho##. This happens when ##\frac{dr}{d\theta}=k(-\sin\theta)=0## which is for Pi and 0.

    Now, help me interpret the acceleration expression I derived above to which I'm expected to provide an interpretation. Also, is my procedure correct?

    Thanks for reading this tedious thing.
     
  2. jcsd
  3. Aug 25, 2015 #2
    Let's take a step back. You kinda have the right idea.

    What is the general equation for the velocity of a particle in cylindrical coordinates in terms of dr/dt and dθ/dt?
    What is the magnitude of this vector?
    For your problem, given the shape of the path, what is the magnitude of the vector (in terms of dθ/dt, but not dr/dt)? (This must be equal to the constant v).

    Chet
     
  4. Aug 25, 2015 #3
    So the velocity vector is ##r' \hat{r} + r \theta ' \hat{\theta}##.

    This would reduce to ##k\theta'(-sin\theta)\hat{r}+k(1+cos\theta)\theta'\hat{\theta}## which, because of its constant magnitude we conclude ##k^2\theta'^2(1+cos\theta)^2+k^2\theta'^2sin^2\theta=v^2##.

    This looks similar to what I found as a term for the radial acceleration but its not exactly the same. Was I supposed to see something important here?
     
  5. Aug 25, 2015 #4
    Yes. You can solve this for dθ/dt.
     
  6. Aug 25, 2015 #5
    ##\theta'=\sqrt{\frac{v^2}{k^2[2+2cos\theta]}}##.
     
  7. Aug 25, 2015 #6
    Am I supposed to solve this DE ? :woot:
     
  8. Aug 25, 2015 #7
    Or should I plug it into my expression?
     
  9. Aug 25, 2015 #8
    Since ##\vec{r}''\cdot \hat{r}=\frac{d}{dt}(-k\theta'sin\theta)-k(1+cos\theta)\theta'^2=\frac{d}{dt}(\vec{v}\cdot\hat{r})-k(1+cos\theta)\theta'^2## I'm starting to think the interpretation is simply that the first term is a radial acceleration by virtue of my change in radial velocity, and the other is a radial acceleration given by the curvature of my curve.
     
  10. Aug 25, 2015 #9
    Where this isn't exactly curvature, but a type of curvature that is seen from the origin, if that even makes sense.
     
  11. Aug 25, 2015 #10
    Well, you are going to need to get r''. First get r' in terms of θ', then substitute your equation in post #5 for θ'. Then take the derivative of the resulting equation with respect to t to get r''. The objective it to express r''-r(θ')2 exclusively in terms of functions of θ.

    Chet
     
  12. Aug 25, 2015 #11
    I've got ##\frac{v^2}{k}(\frac{sin^2\theta}{2+2cos\theta}-\frac{cos\theta}{2+2cos\theta}-1/2]##
     
  13. Aug 25, 2015 #12
    Out of curiosity, what made you think making everything as a function of ##\theta## would be helpful? I'm not seeing it clearly right now. Thanks.
     
  14. Aug 25, 2015 #13
    I'm not going to check your "arithmetic", but I do recommend reducing the final equation to a common denominator and also factoring out that 2.

    Chet
     
  15. Aug 25, 2015 #14
    Lots of experience.

    Chet
     
  16. Aug 25, 2015 #15
    ##\frac{v^2}{2k}(\frac{sin^2\theta-2cos\theta-1}{1+cos\theta})##
     
  17. Aug 25, 2015 #16
    I have a feeling this is wrong since it diverges at ##\theta=pi##
     
  18. Aug 25, 2015 #17
    Hey Chet I've done it all over again and got ##\frac{-3v^2}{4k}## looks interesting.... Any ideas?
     
  19. Aug 25, 2015 #18
    Could this be it? haha
     
  20. Aug 25, 2015 #19
    Chet
    Sure, if you did the algebra correctly. Try doing it a third time and see what you get.

    By the way, the centripetal acceleration is equal to the component of the acceleration vector perpendicular to the velocity vector. Do you know how to determine that?

    Chet
     
  21. Aug 25, 2015 #20
    I think I do. By the way, I think I successfully solved that part in my original post, don't you think? Namely I just found the normal vector in the frenet-serret sense and solved when that would become purely radial and thus the centripetal acceleration would then be what I found as an expression previously. That works doesn't it?
     
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